A sample of argon gas occupies a volume of 1.99 L at -41.3 oC at 1.14 atm. What would its Celsius temperature be at 2.01 atm if its volume decreases to 1.54 L?
To solve this problem, we can use the combined gas law equation:
(P1V1) / T1 = (P2V2) / T2
Where:
P1 = 1.14 atm
V1 = 1.99 L
T1 = -41.3 oC = 231.85 K
P2 = 2.01 atm
V2 = 1.54 L
T2 = unknown
Now we can plug in the values and solve for T2:
(1.14 atm * 1.99 L) / 231.85 K = (2.01 atm * 1.54 L) / T2
2.27 / 231.85 = 3.09 / T2
T2 = (231.85 * 3.09) / 2.27
T2 = 315.67 K
To convert the temperature from Kelvin to Celsius, we subtract 273.15:
T2 = 315.67 K - 273.15
T2 = 42.52 oC
Therefore, the Celsius temperature of the argon gas would be 42.52 oC at 2.01 atm with a volume of 1.54 L.